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CS330: PROBLEMS FOR MIDTERMAPRIL 22, 2011
2.18 We look at a comparison-based algorithm for a search in a
sorted array as a binary tree in which a
path from the root to a leaf represents a run of the algorithm.
At every node a comparison takes placeand, according to its result,
a new comparison is performed. A leaf of the tree represents an
output of thealgorithm: the index of the element x that we are
searching or it says that the element x does not appearin the
array. All indices 1, . . . , n must appear as leaves or the
algorithm will fail when x is at one of themissing indices,
implying that the tree must have at least n leaves, and its depth
must be (log n), whichmeans that in the worst case it must perform
at least (log n) comparisons.
2.30 (a) In order to find for which , 2, . . . , 6 are distinct
modulo 7, we start by checking these powersfor = 2, 3, . . . (Note
that = 1 is trivial so we skip it)
For = 2:
= 2 2 (mod 7)2 = 22 = 4 4 (mod 7)3 = 23 = 8 1 (mod 7)4 = 24 = 16
2 (mod 7)
Since 4 = 2 (mod 7), 2 does not satisfy our conditions.
For = 3:
= 3 3 (mod 7)2 = 32 = 9 2 (mod 7)3 = 33 = 27 6 (mod 7)4 = 34 =
81 4 (mod 7)5 = 35 = 243 5 (mod 7)6 = 36 = 729 1 (mod 7)
It is easy to see that + 2 + 3 + 4 + 5 + 6 = 3 + 2 + 6 + 4 + 5 +
1 = 21 = 0 (mod 7)
(b) The matrix M6() becomes
M6(3) =
1 1 1 1 1 11 3 2 6 4 51 2 4 1 2 41 6 1 6 1 61 4 2 1 4 21 5 4 6 2
3
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and then
FT
011152
= M6(3)
011152
=
1 1 1 1 1 11 3 2 6 4 51 2 4 1 2 41 6 1 6 1 61 4 2 1 4 21 5 4 6 2
3
011152
=
364233
(c) The inverse ofM6() is
M16
(3) = 6
1 1 1 1 1 11 5 4 6 2 31 4 2 1 4 21 6 1 6 1 61 2 4 1 2 41 3 2 6 4
5
Note that the entries in M16
(3) are obtained by taking the multiplicative inverse of the
coresponding entryin M6(3) in modular arithmetic: e.g. 53 = 15 = 1
(mod 7), which means that 3 is at (2,6) and (6,2) positions
in M16 (3) (5 is at these positions in M6(3)).
6 in front of ( ) is1
6(this is
1
nfrom our lectures)
(d) 1 +x+x2 = (1, 1, 1, 0, 0, 0) and 1 + 2x+x3 = (1, 2, 0, 1, 0,
0) = (6, 2, 0, 1, 0, 0). Again all here is done(mod 7), so -1
becomes 6 since the cycles for (mod 7) are. . . , (7,6,5,4,3,2,1),
(0, 1, 2, 3, 4, 5, 6), (7, 8, 9, 10, 11, 12, 13), . . .from where
one can see that for example -1=6=13=-8.
Now
FT
1
11000
= M6(3)
1
11000
=
1 1 1 1 1 1
1 3 2 6 4 51 2 4 1 2 41 6 1 6 1 61 4 2 1 4 21 5 4 6 2 3
1
11000
=
3
60103
and
FT
62010
0
= M6(3)
62010
0
=
1 1 1 1 1 11 3 2 6 4 51 2 4 1 2 41 6 1 6 1 61 4 2 1 4 2
1 5 4 6 2 3
62010
0
=
24431
1
Multiply the obtained FTs componentwise as follows:
FT
111000
FT
620100
=
360103
244311
=
630303
and take the FT inverse of the last vector:2
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April 22, 2011
M16
(3)
630303
= 6
1 1 1 1 1 11 5 4 6 2 31 4 2 1 4 21 6 1 6 1 61 2 4 1 2 41 3 2 6 4
5
630303
= 6
166466
=
611311
which implies that(1 + x + x2)(1 + 2x + x3) = 6 + x + x2 + 3x3 +
x4 + x5 = 1 + x + x2 + 3x3 + x4 + x5
(use 1 = 6 (mod 7))
3.8. (a) Let S0 = 10, S1 = 7, S2 = 4 be the sizes of the
corresponding containers, and let ai be the actualcontents of the
ith container. Let G = (V,E) be a directed graph, in which the
nodes are triples of numbers(a0, a1, a2). It must hold 0 ai Si fori
= 0, 1, 2 and at any given node a0 + a1 + a2 = 11
An edge between two nodes (a0, a1, a2) and (b0, b1, b2) exists
if the following 2 conditions are satisfied:
the two nodes differ in exactly two coordinates, and the third
one is the same in both. if i, j are the coordinates they differ
in, then either ai = 0 or aj = 0 or ai = Si or aj = Sj .
The question is whether there exists a path between the nodes
(0, 7, 4) and (#, 2,#) or (#,#, 2) where #stands for any allowed
value of the corresponding coordinate.
(b) Use DFS on the graph described in (a), starting from node
(0, 7, 4) with an additional line of code thathalts and answers yes
if one of the desired nodes is reached and no if the connected
component of thestarting node is exhausted an no desired vertex is
reached.
(c) (0, 7, 4) (4, 7, 0) (10, 1, 0) (6, 1, 4) (6, 5, 0) (2, 5, 4)
(2, 7, 2)
3.24 First linearize the DAG, and note that since in a
linearized DAG edges can go only in increasing di-rection we can
label its nodes by their position in the linearized order. To check
if there is a directed paththat touches every vertex exactly once
it is enough to check if the DAG has an edge (i,i+1) for every pair
ofconsecutive nodes i and i+1 (i and i+1 being their label
(position) in the linearized order.).
Linerization of DAGs takes linear time (see page 101 in DPV),
and checking the edges for the pairs of con-secutive nodes also
takes linear time, which implies that the total running time is
also linear.
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